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Algebra · real student question

Solve the inequality (75% + 85%(2000 - x))/2000 >= 83% for x.

Question

Solve for xx:

75%+85% (2000−x)2000 ≥ 83%\frac{75\%+85\%\,(2000-x)}{2000}\ \ge\ 83\%

Step-by-step solution

  1. Convert every percentage to a decimal first. Mixing %\% symbols with algebra is the fastest route to an error, so replace them once and for all: 75%=0.7575\%=0.75, 85%=0.8585\%=0.85, 83%=0.8383\%=0.83.

    0.75+0.85(2000−x)2000 ≥ 0.83\frac{0.75+0.85(2000-x)}{2000}\ \ge\ 0.83

  2. Expand the bracket in the numerator.

    0.85(2000−x)=1700−0.85x  ⟹  1700.75−0.85x2000 ≥ 0.830.85(2000-x)=1700-0.85x\;\Longrightarrow\;\frac{1700.75-0.85x}{2000}\ \ge\ 0.83

  3. Clear the denominator. Multiplying both sides by 20002000 is safe without flipping the sign, because 2000>02000>0:

    1700.75−0.85x ≥ 16601700.75-0.85x\ \ge\ 1660

  4. Isolate the xx term. Subtracting 1700.751700.75 from both sides:

    −0.85x ≥ −40.75-0.85x\ \ge\ -40.75

  5. Divide by −0.85-0.85 and flip the inequality. This is the one step where the direction changes, because dividing by a negative number reverses the order:

    x ≤ −40.75−0.85=40.750.85=47.941176…x\ \le\ \frac{-40.75}{-0.85}=\frac{40.75}{0.85}=47.941176\ldots

  6. Check a value on each side. At x=47x=47: 1700.75−39.952000=0.83040≥0.83\tfrac{1700.75-39.95}{2000}=0.83040\ge0.83 ✓. At x=48x=48: 1700.75−40.82000=0.829975<0.83\tfrac{1700.75-40.8}{2000}=0.829975<0.83 ✗ - so the boundary really does sit between 4747 and 4848, and if xx must be a whole number the answer is x≤47x\le47.

Answer

x≤40.750.85=47.941176…(whole numbers: x≤47)x\le\frac{40.75}{0.85}=47.941176\ldots\quad(\text{whole numbers: }x\le 47)

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